Re: Worldmap mercator projection - Latitude to Y calculation.
"J. P. Gilliver" <[email protected]> Thu, 15 Jan 2026 17:25:26 +0000
| Newsgroups | alt.comp.os.windows-xp,alt.windows7.general,alt.comp.os.windows-10 |
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| Organization | 255 software |
| Message-ID | <[email protected]> |
On 2026/1/14 8:24:17, R.Wieser wrote: > Hello all, > > I have a world map using the Mercator projection, and would like to plot > some stuff on it using Latitude and Longitude. > > The problem is that I can't seem to get the formule I found to spit out the > correct value for Y. > > -- The map : > > https://upload.wikimedia.org/wikipedia/commons/thumb/7/73/Mercator_projection_Square.JPG/250px-Mercator_projection_Square.JPG > > -- The formula : > > Y = ln( tan(latitude) + sec(latitude) ) > > where "ln(...)" is log(...) / log(10) > > ... at least, that is what I could google about it. Rather than Googling the formula, my first thought was to go back to first principles: as the name implies, a map projection can be thought of as being created by shining a light through the earth (globe) onto a sheet of paper - either flat and fixed to the globe at one point, or - more commonly - a cylinder wrapped round the globe, touching at one circle (often the equator), and then unrolled. Once this is realised, basic geometry should make calculation of the co-ordinates fairly simple. Unfortunately, to do this, one needs to know where the nominal light source is. And - despite it being a _long_ article - I can't find a statement of this in the Wikipedia article about the Mercator projection - other that that - I _think_ - it _isn't_ the centre of the earth (so-called "radial"), though is often thought to be (certainly Google's AI thinks it is). If anyone _can_ find out where the light source point is for the Mercator projection, I'd love to know! (And it would answer Rudy's question.) (One other position for the light source I remember from last time I looked into this - which was probably over 40 years ago! - is on the opposite surface of the globe to the projection point (i. e. sort of tracking round opposite the map "printing"); I don't think it's that, though, as that would show the poles, though still distorted. If it _is_ light-source-at-centre, then the Y co-ordinate would just be the tangent of the latitude (scaled appropriately for the map size). Even if it isn't, this _may_ be close enough - try a few places. -- J. P. Gilliver. UMRA: 1960/<1985 MB++G()ALIS-Ch++(p)Ar++T+H+Sh0!:`)DNAf It has been my experience that folks who have no vices have very few virtues -- Abraham Lincoln, quoted by Mark Lloyd in alt.windows7.general 2018-12-27